Problem: Graph this system of equations and solve. $10x-10y = -50$ $4x+5y = -20$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Solution: Convert the first equation, $10x-10y = -50$ , to slope-intercept form. $y = x + 5$ The y-intercept for the first equation is $5$ , so the first line must pass through the point $(0, 5)$ The slope for the first equation is $1$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move up You must also move $1$ position to the right. $1$ position to the right. $1$ position up from $(0, 5)$ is $(1, 6)$ Graph the blue line so it passes through $(0, 5)$ and $(1, 6)$ Convert the second equation, $4x+5y = -20$ , to slope-intercept form. $y = -\dfrac{4}{5} x - 4$ The y-intercept for the second equation is $-4$ , so the second line must pass through the point $(0, -4)$ The slope for the second equation is $-\dfrac{4}{5}$ . Remember that the slope tells you rise over run. So in this case for every $4$ positions you move down (because it's negative) $5$ positions to the right. $4$ positions down from $(0, -4)$ is $(5, -8)$ Graph the green line so it passes through $(0, -4)$ and $(5, -8)$ The solution is the point where the two lines intersect. The lines intersect at $(-5, 0)$.